Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $q = \dfrac{z - 3}{z^2 - 12z + 27} \times \dfrac{-5z^2 + 45z}{2z + 14} $
Explanation: First factor the quadratic. $q = \dfrac{z - 3}{(z - 9)(z - 3)} \times \dfrac{-5z^2 + 45z}{2z + 14} $ Then factor out any other terms. $q = \dfrac{z - 3}{(z - 9)(z - 3)} \times \dfrac{-5z(z - 9)}{2(z + 7)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (z - 3) \times -5z(z - 9) } { (z - 9)(z - 3) \times 2(z + 7) } $ $q = \dfrac{ -5z(z - 3)(z - 9)}{ 2(z - 9)(z - 3)(z + 7)} $ Notice that $(z - 3)$ and $(z - 9)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -5z(z - 3)\cancel{(z - 9)}}{ 2\cancel{(z - 9)}(z - 3)(z + 7)} $ We are dividing by $z - 9$ , so $z - 9 \neq 0$ Therefore, $z \neq 9$ $q = \dfrac{ -5z\cancel{(z - 3)}\cancel{(z - 9)}}{ 2\cancel{(z - 9)}\cancel{(z - 3)}(z + 7)} $ We are dividing by $z - 3$ , so $z - 3 \neq 0$ Therefore, $z \neq 3$ $q = \dfrac{-5z}{2(z + 7)} ; \space z \neq 9 ; \space z \neq 3 $